Suppose the contrary. Then \(\lambda \left|G_{x,B}\right|< \left[ G_{x}:X_{x}\right] \leqslant \left|\mathrm {Out(}X\mathrm {)}\right|\) , as \(\left[ G_{x}:H_{x}\right] \)  divides \(\left|\mathrm {Out(}X\mathrm {)}\right|\) . Also, \(\lambda \left|G_{x,B}\right|=\rho \theta \) , where \(\rho =\left(\Phi _{ef}^{\ast }(p),\lambda \left|G_{x,B}\right|\right) \)  and \(\theta \geqslant 1\) . Since \(\rho >1\) , \(\rho =u_{1}^{m_{1}}\cdots u_{\ell }^{m_{\ell }}\)  with \(u_{i}\)  prime for \(i=1,...,\ell \) , and \(u_{i}\ne u_{j}\) 
for \(i\ne j\) . Furthermore, \(u_{i}\equiv 1 \pmod {ef}\)  by [1]}, Proposition
5.2.15.(i), and hence \(u_{i}\geqslant 1+ef\) . Therefore, either \(\rho \geqslant 3\left(1+ef\right) \)  (actually, \(\rho >6\left( 1+6f\right) \)  for \(P\Omega _{8}^{+}(q)\)  by Remark REF ), or \(\rho =1+ef\)  is a prime. Consequently, \(\left|G_{x,B}\right|\geqslant \mu (1+ef) \theta \) , where \(\mu \geqslant 3\) , or \(\mu =1\)  and \(1+ef\)  is a prime.
Assume that \(X\)  isomorphic to one of the groups \(PSL_{n}(q)\) 
or \(PSU_{n}(q^{1/2})\) . Then \(\left|\mathrm {Out(}X\mathrm {)}\right|\leqslant 2nf\)  by [1]}, Table 5.1.A, and hence \(\mu \left( 1+ef\right) \theta \leqslant \lambda \left|G_{x,B}\right|\leqslant 2nf\) . Then \(\mu = 1\) , \(\rho =1+ef\)  is a prime and either \(e=n\)  and \(\theta =1\) , or \(e=n-1\) , \(X\cong PSU_{n}(q^{1/2})\) , with \(n\)  even, and \(\theta \leqslant 2\) .
If \(\theta =1\) , then \(\lambda \left|G_{x,B}\right|\)  is odd
and hence \(\rho =\lambda \left|G_{x,B}\right|\leqslant \left[ G_{x}:H_{x}\right] <1+ef\)  by Lemma REF (1). However, this contradicts \(\rho = 1+ef\) .
If \(\theta =2\) , then \(X\cong PSU_{n}(q^{1/2})\) , with \(n\)  even. It results \(2 \rho = \lambda \left|G_{x,B}\right|\leqslant \left[ G_{x}:H_{x}\right] <1+ef\) 
by Lemma REF (2), and we again reach a contradiction.
Assume that \(X\)  isomorphic to one of the groups \(PSp_{n}(q)\)  or \(P\Omega _{n}(q)\) . Then \(\left|\mathrm {Out(}X\mathrm {)}\right|\leqslant 2f\)  by
[1]}, Table 5.1.A, and hence \(\theta \left( 1+ef\right) \leqslant \lambda \left|G_{x,B}\right|\leqslant 2f\) , whereas \(e=n-2\geqslant 2\)  by Remark REF .
Assume that \(X\)  isomorphic to \(P\Omega _{n}^{\varepsilon }(q)\) , with \(\varepsilon =\pm \)  and \((n,\varepsilon )\ne (8,+)\) . Then \(\left|\mathrm {Out(}X\mathrm {)}\right|\leqslant 8f\)  by [1]}, Table 5.1.A, and hence \(\mu \left( 1+ef\right) \theta = \lambda \left|G_{x,B}\right|\leqslant 8f\) . Then \(\mu =\theta =1\) , \(n=6\)  and \(\rho =1+ef\)  is a prime, and hence \(\lambda \left|G_{x,B}\right|=1+ef \) , whereas \(\lambda \left|G_{x,B}\right|<1+ef \)  by Lemma REF .
Finally, assume that \(X\cong P\Omega _{8}^{+}(q)\) . Then \(\lambda \left|G_{x,B}\right|\leqslant \left|\mathrm {Out(}X\mathrm {)}\right|\leqslant 24f\)  and hence \(\lambda \left|G_{x,B}\right|=\theta \rho \) , with \(\theta \leqslant 3\)  and \(\rho =1+6f\)  a prime. Moreover, either \(\theta =1,3\)  and \(\lambda \left|G_{x,B}\right|\)  is odd, or \(\theta =2\)  and \(\lambda \left|G_{x,B}\right|\equiv 2\pmod {4}\) . Then \(\lambda \left|G_{x,B}\right|\leqslant \left[ G_{x}:H_{x}\right] <\rho \)  by Lemma REF (2)–(3), and we again reach a contradiction. This completes the proof.
